A gambler playing roulette has a superstition that he must place bets for exactly 10 spins of a roulette wheel (American wheel with 00). He has a bankroll of $300 and the table limits are minimum $1 and maximum $150, with $1 increments.

Can you find a strategy which guarantees he will be able to place ten bets and will have at least a 96% chance of making a profit (finishing bankroll of at least $301)?

Consider that we only need to find a 4% solution where he loses money.  Since he must place 10 bets, a minimum of $10 must be wagered.  Therefore, he must win enough  in one bet to cover losing the other nine.

The probability of a win on an "even" wager in roulette is 36/38, or roughly 94.737%.  He starts off by betting one dollar on the first five rolls.  It matters not whether he wins or not.  Let's assume he loses all five.  On the sixth roll, he wagers $10.  If he wins, he can afford to lose a dollar bet on the last four rolls which nets him a $1 profit.

If he loses roll 6, then he bets $19 on roll 7.  This covers his $10 loss on the previous roll plus $1 losses on the other eight and nets him a $1 profit.

If he loses roll 7, he bets $37 on roll 8.  This will cover his $19 loss, his $10 loss, plus seven $1 losses, netting him a profit of $1.

If he loses roll 8, he bets $73 on roll nine which after covering his bets will net him a profit of $1.

If he loses roll 9, he bets $145 on the final roll.  If he wins, he covers the previous $144 in loses, netting him a profit of $1.

The only way he can come out behind is if he loses the last 5 rolls in a row.  The probability of this occurring is (20/38)^5 or 4%.  This means the probability of it not occurring is 96%.

Posted by hoodat on 2007-04-05 14:41:28